In this section, without specific mention, all the domains, functions, etc. are asregular as we want.
1.1 Graph
Suppose Ω is a domain in R3, and u : Ω→ R is a function, and Gu is the graphof u.
Then the tangent space to Gu is spanned by two vectors ∂x = (1, 0, ux) and∂y = (0, 1, uy).
The normal vector is defined as
N =(−ux,−uy, 1)√
1 + |∇u|2
The area of Gu is by definition∫Ω
√1 + |∇u|2dxdy
Question When dos Gu have the least area for its boundary?
1.2 Variation
We consider the variation of this integral. Consider u + tη, where η : Ω → R,and vanishes on boundary. Then we define:
A(t) =
∫Ω
√1 + |∇u|2 + 2t〈∇u,∇η〉+ t2|∇η|2
Take derivative we have
A′(t) =
∫Ω
〈∇u,∇η〉√1 + |∇u∇|2
=
∫Ω
ηdiv(∇u√
1 + |∇u|2)
It should be 0 for any η, if Gu is minimal. So we get the minimal surfaceequation (MSE):
div(∇u√
1 + |∇u|2)
We call the solution to this equation is minimal surface.
2
remark 1.1. When |∇u| is bounded, this equation is uniformly elliptic equation,so we can use some classical theory;
When |∇u| → 0, Then this equation is in some sense linearization of zerosolution. Which means the equation is close to Laplacian equation.
remark 1.2. The solution to the MSE includes the planes: u = ax+ by + c
1.3 Area Minimizing Property of Gu
If Gu is area minimizing, then it satisfies the MSE. How about the conversecase? Note if Gu satisfies MSE, it is just a ”critical solution”, not a ”minimalsolution”
lemma 1.1. If Gu is minimal and Σ ⊂ Ω×R with ∂Σ = ∂Gu, then Area(Gu) ≤Area(Σ)
So the lemma tells us ”critical is minimal”
Proof. Recall N =(−ux,−uy,1)√
1+|∇u|2is the normal vector on Gu. Extend it to Ω×R,
which is independent of z.By MSE,
divR3(N) = divR2(−∇u√
1 + |∇u|2) = 0
By stokes theorem, we have
∫Gu
N · normalGu −∫
Σ
N · normalΣ =
∫volume between Ω and Σ
divR3(N) = 0
⇒ Area(Gu) =
∫Σ
N · normalΣ ≤ Area(Σ)
Note that this lemma is not true if Σ * Ω× R. To see this, see picture 1.However, this is true if Ω is convex. We can then replace Σ − Ω × R by its
projection to the boundary, hits is area non-increasing. (WHY?)
remark 1.3. The idea of the proof of the lemma is called calibration.In general, for a p-dimensional hypersurface, we define the calibrate form ω
as a p-form, which equal to the volume form on the hypersurface, and for q anyp-vector, we have ω(q) ≤ 1
Corollary 1.1. Suppose Gu is minimal, Dr ⊂ Ω is a disk of radius r, thenArea(Br ∩ Gu) ≤ 2πr2, here Br is the ball with radius r, with center has thesame x, y coordinates with the center of Dr
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Proof. We compare the minimal surface with sphere. By Sard theorem, for εarbitrary small, we can find a sphere Br−ε′ which intersects Gu transversally,where ε′ < ε. Then Gu split this sphere into at least two parts. Choose the partwhich has less area, and use the lemma above, and let ε→ 0 we then prove thetheorem.
2 Geometry of submanifold
2.1 Notation and Definition Review
Suppose Σ is a hypersurface, in the ambient space M . Let e1, e2 · · · , en is theorthonormal frame of Σ
Let V is a vector along Σ, then we can divided it into two parts
V = V > + V ⊥
Which are the tangential part and the vertical part (sometimes we write it asV N ).
Suppose X,Y are vector fields along Σ, both are tangential, then
∇XY = ∇>XY +∇⊥XY
The first part is called the induced connection on Σ, and the second part iscalled the second fundamental forms, which denoted by A(X,Y )
We define Mean Curvature H = tr(A) =n∑i=1
A(ei, ei) and we define the
norm of A by |A|2 =n∑
i,j=1
|A(ei, ej)|2
We define Divergence along Σ by divΣ(X) =n∑i=1
〈∇eiX, ei〉
A fact:
divΣ(Y N ) =
n∑i=1
〈∇eiY N , ei〉 =
n∑i=1
[ei〈Y N , ei〉 − 〈Y N ,∇eiei〉]
= −〈Y N ,n∑i=1
∇eiei〉 = −〈Y N ,n∑i=1
∇⊥eiei〉 = −〈Y,H〉
Which means calculate the divergence of a vector fields on a submanifold isexactly the same to calculate the inner product of it with mean curvature.
2.2 Variation
Consider F : Σ × (−ε, ε) → Rn+1 is a family of maps parameterized by t, withboundary fixed at any time t. F (x, 0) is just identity map to the original Σ.Choose a local orthonormal frame ∂i = ∂xi , and push it to the frame on theimage denoted by Fi = Fxi .
4
Write gij = 〈Fi, Fj〉 is the metric. Then we can write down the volume attime t as:
vol(t) =
∫Σ
√detgij(t)dx =
∫Σ
√detgij(t)√detgij(0)
dVΣ
Note that for the later part of the above formula, V (x, t) =
√detgij(t)√detgij(0)
is
independent with the coordinates at fixed point. Thus we can choose good localframe to calculate. Choose our coordinates such that gij(x, 0) = δij , whichmeans Fi are orthonormal frames at (x, 0)
vol′(0) =
∫Σ
V ′(x, 0)dVΣ
We calculate V ′:
v′(x, 0) =1
2[detgij(x, t)]
′|t=0
=1
2Tr(g′ij(x, 0)) =
n∑i=1
〈Fxi,t, Fxi〉 =∑i
〈∇FiFt, Fxi〉 = divΣ(Ft)
So if Σ is minimal, we must have divΣ(Ft) = 0 for any variation V withcompact support.
In particular, if Ft is normal, then by previous formula, we have
divΣ(Ft) = −〈Ft, H〉Thus Σ is critical point of area is equivalent to H = 0.
2.3 Harmonic Coordinate
lemma 2.1. Coordinate functions are harmonic on minimal Σ
This is interesting, because it tells us that an extrinsic function (coordinatefunctions are natural for the ambient space, but not natural for the minimalsurface).
Proof. We calculate the Laplacian of coordinate functions explicitly:
∇Σxi = ∇>xi = ∂>i
∆Σxi = divΣ(∇Σxi) = divΣ(∂i − ∂⊥i )
Note ∂i is a constant vector field, so
= −divΣ(∂⊥i ) = 〈H, ∂⊥i 〉 = 0
Since Σ is minimal.
5
The most basic fact about harmonic functions is the maximal principle. Byusing maximal principle we can get the following corollary:
Corollary 2.1. Given any coordinate function u on Σ, u must has its maximumand minimum on the boundary of Σ.
Moreover, by this fact, if Σ is minimal, then Σ ⊂ conv(∂Σ), where conv(X)is the convex hull of set X.
Proof. Suppose there is a point x on Σ which doesn’t lies in conv∂Σ. Then thereis a hyperplane such that x belongs to it but it doesn’t intersect with conv∂Σ.To see this first note ∂Σ is compact, that means conv∂Σ is closed. Then we canchoose the plane orthogonal to the project segment from x to conv∂Σ.
Then we choose the coordinate function which is the height function verticalto this hyperplane. By the lemma above we get a contradiction.
Let’s see an application in topology of minimal submanifold. We just seethe simplest case:
Corollary 2.2. (Monotonicity of Topology) Suppose Σ ⊂ R3 is minimal andsimply connected, compact with boundary ∂Σ. Suppose BR is a ball disjointfrom ∂Σ, then BR ∩ Σ is a union of disks.
Proof. Since Σ is simply connected, then for any closed curve γ ⊂ BR ∩ Σ, γmust bound a disk Γ ⊂ Σ.
By previous corollary, this minimal surface Γ lies inside conv(γ) ⊂ BR.Which implies the corollary.
3 Scale Invariant Monotonicity Quantity
Next, we want to find a scale invariant monotonicity quantity. This is an ideain geometric problem and differential equation theory.
Note scale invariant is important, because that helps us consider the problemreally geometrically (But not depends on the parameters).
3.1 Volume Density
Consider Σk ⊂ Rn is a minimal submanifold. Let us consider the quantity:
ΘR =vol(BR ∩ Σ)
Rk
(Compare to Bishop-Gromov theorem)It is an obvious scale invariant.
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3.2 Baby Case
Let us first consider the most simple case: Σ = Rk is the hyperplane.Then ∆Rk |x|2 = 2k, so we have:
V (R) = vol(BR ∩ Rk) =1
2k
∫BR
∆Rk|x|2
=1
2k
∫∂BR
2〈(x1, x2, · · · , xk, 0, · · · , 0), outer normal to BR)〉 =1
2k
∫∂BR
R
So we have
V =R
kV ′ ⇒ log(
V
Rk) = 0
So we find the quantity is nondecreasing (in fact, a constant).
3.3 Coarea Formula
Hope I can fulfill this part in few weeks
Theorem 3.1. ∫f≤t
h|∇f | =∫ t
−∞
∫f=s
hds
3.4 General Case
let V (R) = vol(BR ∩ Σ). We have
V (R) =
∫|x|≤R
|∇Σ|x||−1|∇Σ|x|| =∫ R
0
∫∂BR∩Σ
|∇Σ|x||−1
so
V ′(R) =
∫∂BR∩Σ
|∇Σ|x||−1
Note, since Σk is minimal, we have
∆Σ|x|2 = divΣ(2x>)
(This step is because ∇Σ|x|2 = ∇>|x|2 = 2x>)
= 2divΣ(x− x>) = 2divΣ(x) = 2k
By Stokes theorem, we have
V =1
2k
∫∂BR∩Σ
∇>|x|2 · normal =R
k
∫∂BR∩Σ
|∇>|x||
7
In the class Bill gave two proof of monotonicity.Method 1: just note since |∇>|x|| ≤ 1⇒ V ′ ≥
∫∂BR∩Σ
1Also notice
V ≤ R
k
∫∂BR∩Σ
1
⇒ V ′
V≥ k
R⇒ V
Rk≥ 0
We can see when we have equality. Equality holds when:
|∇>|x|| = 1
That means x always tangent to Σ. Thus, Σ is invariant under dilation,means Σ is a minimal cone.
Method 2: We directly calculate the derivative.
(V
Rk)′ =
V ′
Rk− k V
Rk+1=
∫∂BR∩Σ
[|∇Σ|x||−1
Rk− |∇Σ|x||
Rk]
=
∫∂BR∩Σ
|x|−k−1 |x⊥|2
|x>|
⇒ (V
Rk)′ =
∫∂BR∩Σ
|x|−k−2|x⊥|2(|x||x>|
))
Here we want to use coarea formula (Note ∇Σ|x| = x>
|x| )
Then we integrate both part, by coarea formula, we have
V (t)
tk− V (s)
sk=
∫Bt∩Σ−Bs
|x⊥|2
|x|k+2≥ 0
remark 3.1. We have monotonicity about the quantity Θ(R). Then we canwell define the quantity Density at point 0 by
Θ0 = limR→0
V (R)
Rk≥ 0
(Or we divide it by a dimensional constant)
We can also prove the monotonicity of the quantity in some special case ina simpler way:
Let σ = ∂BR ∩ Σ, and let C is the cone on σ. Then we have
vol(C) =R
k|σ| ≥ vol(BR ∩ Σ)
⇒ R
k|vol(∂BR ∩ Σ)| ≥ vol(BR ∩ Σ)
This is again the same differential equation we get before.
8
4 Density
Intuitively, density describe the local behavior of the point. Since every smoothsurface Σ is locally R2, we can image at point x0 ∈ Σ we must have Θ(x0) = π
This is proved by Allard:
Theorem 4.1. For Σ2 ⊂ R3 minimal, then we have:1. Θ = π at all smooth points;2. Θ > π + ε at all singular points, for some ε > 0
remark 4.1. This type of ”ε-regularity” theorems also appear in many otherproblems. The main spirit of this type of theorem is that if the scale invari-ant monotonicity quantity close to its minimal value, then the underlying spaceshould be smooth.
We come back to Allard theorem later. Now let’s discuss a property of Θ.Let Θx denote the density at point x:
lemma 4.1. Θy is upper semi-continuous in y. (i.e., if yi → y, then Θy ≥lim sup Θyi)
The by Allard theorem, we can get a corollary: singular points are closed.
Proof. Choose r small enough such that
|vol(Br(y) ∩ Σ)
rk−Θy| < ε
Then if we choose yi close to y, we have:
Θyi ≤vol(Br−|y−yi|(yi) ∩ Σ)
(r − |y − yi|)k≤ (Br(y) ∩ Σ)
(r − |y − yi|)k
Assume |y − yi < εr, then we have
Θyi ≤(Br(y) ∩ Σ)
(1− ε)krk
Let ε→ 0 we get the required estimate.
4.1 Mean Value Inequality
We define density to be an quantity related to area, which is integral over themanifold with integrate function 1. How about a general function?
Suppose Σk ⊂ Rn is minimal as before. f is a function over ΣLet I(R) =
∫Br∩Σ. If f ≡ 1 this is just volume as before.
Theorem 4.2.
I(t)
tk− I(s)
sk=
∫Bt−Bs
f|x⊥|2
|x|2+k+
1
2
∫ t
s
r−k−1[
∫Br∩Σ
(r2 − |x|2)∆Σf ]
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Corollary 4.1. If ∆Σf ≥ 0, then for t ≥ s
I(t)
tk≥ I(s)
sk≥ lims→0
I(s)
sk
If Σ is smooth, by Allard theorem the quantity satisfies:
= f(0)vol(B1 ⊂ Rk)
Then we have
f(0) ≤∫Bt∩Σ
f
vol(Bt ⊂ Rk)
Proof. Let’s prove mean value inequality.Just as previous proof of density monotonicity, recall ∆Σ|x|2 = 2k. Then we
Then apply the coarea formula we get what we want.
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Corollary 4.2. Suppose ∆Σf ≥ −λf and f ≥ 0, and 0 ∈ Σ. Then we have
f(0) ≤ eλ2∫B1∩Σ
f
vol(B1 ⊂ Rk)
Proof. Note if λ = 0 then is just subharmonic case as before.Let g(t) = t−k
∫Bt∩Σ
f for t ≤ 1Then we have
g′(t) ≥ 1
2t−k−1
∫Bt
(t2 − |x|2)(−λf) ≥ −λ2t−k+1
∫Bt
f ≥ λ
2tg ≥ −λ
2g
By the differential inequality we can get the result.
4.2 ∆f ≥ −f 2?
We have discussed the mean value property for subharmonic function and sub-eigenvalue functions. How about f satisfies ∆f ≥ −f2? Does the previousthings work for this f?
The answer if NO. In fact, we can find f such that ∆f ≥ −f2, and∫B1∩Σ
f =
1, but f(0) can be arbitrary large.This is a bad news for minimal surface. In fact, Simon inequality:
∆|A|2 ≥ −2|A|4
Is in this form. Thus, many estimate should first assume the second funda-mental form is bounded.
5 Strong Maximal Principle
Recall the strong maximal principle for elliptic pdes:
Theorem 5.1. Suppose w is a function over a domain Ω, satisfies equation:
div(Aij∂jw) = 0
where Aij(x) = Aji(x) is positive definite smooth function on Ω. Then wcan not have interior minimum value unless w is a constant.
This is the form we want to analysis minimal surface. We want to show thefollowing property of minimal surfaces:
Theorem 5.2. Suppose Σ and Γ are two smooth compact connected minimalsurface and ∂Σ ∩ Γ = ∅, ∂Γ ∩Σ = ∅. If Σ and Γ intersect but do not cross eachother, then they coincide.
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Observe that suppose p is such a point that two minimal surface intersectbut do not cross each other, then locally these two surfaces are graph over thetangent space at p. Thus, we only need to consider the problem under theassumption that Σ and Γ are graph of function u and v.
Proof. set w = u− v. By MSE, we have
div(∇u√
1 + |∇u|2) = div(
∇v√1 + |∇v|2
) = 0
Define F : Rn → Rn such that F (a) = a√1+|a|2
, then we have
div(F (∇u)− F (∇v))
we compute
dFY (Z) =Z√
1 + |Y |2− 〈Y,Z〉Y
(1 + |Y |2)3/2
Then we have
〈dFY (Z), Z〉 =|Z|2√
1 + |Y |2− 〈Y, Z〉2
(1 + |Y |2)3/2> 0
if Z 6= 0So dF is positive definite. Thus
F (∇u)− F (∇v) =
∫ 1
0
dF(∇v+t∇w)(∇w)dt = 0
is an elliptic equation in our previous maximal principle setting. So w sat-isfies maximal principle, which means if w = 0 in the interior some where thenwe must have u = v, which means two surface coincident.
5.1 Rado-Schoen theorem
This section needs a lot of pictures, so I update it later.
6 Second Variation Formula
In 1968, J.Simons gave a fundamental variation formula for minimal surface.Suppose F : Σ× (−ε, ε)→Mn+1 satisfies the following conditions:(1)Σ is minimal and two sided (here two sided means Σ has trivial normal
bundle);(2)F (·, 0) is identity map;(3)Ft has compact support;(4)F>t = 0Then we have the second variation formula:
12
vol′′(0) = −∫
Σ
〈Ft, LFt〉
here L is an operator
L = ∆⊥Σ +RicM (n, n) + |A|2
Note the ∆Σ is the non-positive term, and RicM (n, n)+ |A|2 is non-negativeterm if Ric ≥ 0. This is the requirement always appear in minimal surfacetheory.
In the rest of this section, we will prove the second variation formula.
Proof. Let us choose local coordinates xi on Σ, which induce metric gij(x, t)Define
V (x, t) =
√detgij(x, t)√detgij(x, 0)
We want to calculate the second derivative.By computation of first order derivative, we have
V ′ =1
2V trace(g−1g′)
This can be derived from directed computation. Next we calculate the secondvariation of V :
V ′′(0) =1
2V ′Trace(g−1g′) +
1
2V Trace((g−1)′g′) +
1
2V Trace(g−1g′′)
For the rest of the proof, let us fix a point x and choose a local coordinatesuch that gij(x, 0) = δij . Also note that g−1g = 1, take derivative we get(g−1)′ = −g−1g′g−1
So in this setting, note at time t = 0, we have V = 1, V ′ = 0:
V ′′(0) =1
2Trace((g′)2) +
1
2Trace(g′′)
Since g′ is a symmetric matrix, we have Trace((g′)2) = |g′|2Now we need two lemma:
Note by Stokes theorem and minimality, we have the integral of the last termis zero. Then we finish the proof.
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6.1 Stability
We can a minimal surface is stable is V ′′(0) ≥ 0 for any compact supportedvariation Ft.
Stability involves in many interesting problems in minimal surface theory.For instance: if Σ is stable and 2-sided, then we can write down the secondvariation formula in the following form:∫
Σ
fLf ≤ 0
Here Ft = fn is the variation. This formula is exactly:∫Σ
f∆Σf +RicM (n, n)f2 + |A|2f2 ≤ 0
i.e. ∫Σ
(|A|2 +RicM (n, n))f2 ≤∫
Σ
|∇Σf |2
We can get many result from this solution. For instance, For any ambientmanifold with Ricci curvature greater than 0, if we choose f = 1 over the min-imal closed subsurface, we know that there is no stable 2-sided closed minimalsurfaces in this manifold.
There also many other properties concerning stability. Maybe add some herelater.
6.2 More about Stability
We continue considering properties of stable minimal surface. The startingpoints are always second variational formula. If the operator L is related tosome stable minimal surface, then sometimes we also call L is stable.
lemma 6.3. L is stable if there exists a function u > 0 on Σ such that Lu = 0
Intuitively, this can be viewed from the perspective of spectrum theory.Consider that L is symmetric, then we can diagonalize it to get all eigenvaluesλi and eigenfunctions ui. i.e.
Luj = −λjujby spectrum theory
λ1 ≤ λ2 ≤ · · · → ∞
if we can prove that 0 = λ1 is the lowest eigenvalue, then we done. Note ifu1 > 0, in fact we get λ1 is multiplicity 1. (Why?)
Proof. Take f is a compactly supported function on Σ, then we have
div(f2∇uu
) = 2f〈∇f, ∇uu〉+ f2 ∆u
u− f2 |∇u|2
u2
15
Define
q = |A|2 +RicM (n, n)
Note by Lu = 0, we have
∆u
u= −q
So ∫f2q =
∫−f2 |∇u|2
u2+ 2f〈∇f, ∇u
u〉
=
∫|f∇uu−∇f |2 + |∇f |2 ≤
∫|∇f |2
This means for any f it satisfies the stability condition.
As a corollary, we can show minimal graphs are always stable:
Corollary 6.1. Minimal graphs in R3 are stable.
Proof. let ∂z is the constant vector field point to the z- direction everywhere inR3. Let f = 〈∂z, n〉. Since Σ is a graph, f > 0. so we by the lemma above weonly need to prove that Lf = 0.
Note the ambient space is flat, so we do not need to consider the curvatureterm.
Let us compute L〈∂z, n〉 terms by terms. Let us choose a local geodesiccoordinate xi on the surface.
By Bianchi identity, Aji,i = Aii,j , by minimality of the surface, Aii,j = 0. Bythe selection of local geodesic frame we have at one point x, we have ∇Σ
i ∂j = 0,i.e. ∇i∂j = Aij
So we have:
∆Σ〈∂z, n〉 = −〈∂z, n〉|A|2
Combined with the other term of L we have the result.
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6.3 Parabolic
Definition 6.1. We call a complete non-compact manifold Σ is parabolic ifthere exist a sequence of functions uj with compact support such that:
(1)uj → 1 on compact sets of Σ(2)
∫Σ|∇uj |2 → 0 as j → +∞
Parabolic is a useful property when we analyze stable minimal surfaces.
Theorem 6.1. Suppose Σ is a 2-sided stable parabolic minimal hypersurface ofM , while M has non-negative Ricci curvature.
Then we have Σ is totally geodesic.
Proof. By stability we have the inequality:∫Σ(|A|2 +RicM (n, n))u2
j ≤∫
Σ
|∇uj |2
Note by definition of parabolic, the right hand side will goes to 0 whenj →∞, meanwhile the left hand side will goes to
∫Σ|A|2. Hence A vanishes on
Σ everywhere. Thus, Σ is totally geodesic.
Theorem 6.2. suppose Σ is a complete hypersurface with no boundary, for anyr > 0 Area of Br ≤ Cr2. Then Σ is parabolic.
Before we prove this property, we first see an example: how to show that R2
is a parabolic manifold.Let us construct the functions
uj(x) =
1 on Bej
2− log rj on Be2j −Bej
0 outside
Notice the construction use the Newtonian potential in R2. Then we have∫|∇uj |2 = 2π
j , so R2 is parabolic by definition.Now we prove the theorem:
Proof. Similarly construction. We construct a sequence of functions as:
uj(x) =
1 on Bej
2− log rj on Be2j −Bej
0 outside
Note here the balls are the intrinsic ball over the manifold.Though we still have |∇uj | = 1
jr |∇r| ≤1jr in the non-constant part of the
manifold, we can not just prove the estimate as in the example because now themanifold is no longer homogeneous. However we can estimate the integrationin bands, i.e.:
17
∫Bej+1−Bej
|∇uj |2 ≤1
j2
1
e2jArea(Bej+1)
by the assumption we have∫Bej+1−Bej
|∇uj |2 ≤ Ce2
j2
Then we integrate∫Be2j−Bej
into these pieces, we have∫Be2j−Bej
|∇uj |2 ≤ Ce2
j
So we proved that Σ is parabolic.
Corollary 6.2. (Bernstein Theorem) Any entire minimal graph in R2 is aplane.
Proof. Since the minimal graph satisfies the local minimal property:
Area(Br) ≤ 2πr2
So by the lemma the minimal surface is parabolic. Since it is also 2-sidedand stable, it is totally geodesic so it is a plane.
Finally we show one more property related to parabolic.
lemma 6.4. Suppose Σ is parabolic, and u is a positive superharmonic over Σ,then u is a constant.
Proof. By computation we have:
div(u2j
∇uu
) = 2uj〈∇uj ,∇uu〉+ u2
j
∆u
u− u2
j
|∇u|2
u2
By integration we have:∫u2j
|∇u|2
u2≤
∫2〈∇uj ,
uj∇uu〉12
∫uj|∇u|2
u2+
∫|∇uj |2
Then we have ∫u2j
|∇u|2
u2≤ 4
∫|∇uj |2 → 0
Let j →∞ we have:
|∇u| = 0
So u is constant.
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7 Simons Inequality
In 1968, Simons discovered the important inequality for second fundamentalproof. This is one of the most important technique in minimal surface theory.
Theorem 7.1. Suppose Σ is minimal, then we have
∆ΣA = −|A|2A
From this equality we have:
1
2∆Σ|A|2 = |A|4 + |∇A|2
Hence we have
1
2∆Σ|A|2 ≥ −|A|4
The last inequality we will use in the curvature estimate.Before we prove the equality, let us first recall the Codazzi equation. Codazzi
equation is about the calculation of the second fundamental forms:
Thus by the definition of Riemannian tensor we know that
(∇UA)(V,W ) = (∇VA)(U,W )
So the derivative of the second fundamental forms are fully symmetric.Now we prove Simons inequality. We show the general case:
Theorem 7.2. Consider Σn → Rn+1, here Σ is not necessary minimal. Thenwe have
∆ΣA = −HessH − |A|2A−HA2
Just a caution: note here the different definition of H (the symbol) willinfluence the formulate of the result. We define H such that Sn has H > 0
19
Proof. Let us first accept a lemma here:
Aij,kl −Aij,lk = RΣlkimAmj +RΣ
lkjmAmi
It is easy, just directly computation. So we can compute:
∆A(ej , ej) = Aij,kk
By Codazzi equation:
= Aik,jk = Aik,kj +RkjimAmk +RkjkmAmi
Note the first term is Akk,ij by Codazzi equation, which is the Hessian term.This term vanishes in minimal case. The rest term, by Gauss equation:
Rijkl = AikAjl −AilAjkSo we have
∆ΣA = −HessH − |A|2A−HA2
Then we can naturally get Simons inequality.
Here in fact we reduce the difficulty in general case, since the ambient spaceis flat. If the ambient space is non-flat, then we can get a more general Simonsinequality:
1
2∆Σ|A|2 ≥ −λ|A|4 + |∇A|2
Where λ depends on the curvature and the derivative of the curvature of theambient space.
7.1 Curvature Estimate by Choi and Schoen
Choi and Schoen used Simons inequality showed the following result (we considerthe easiest case):
Theorem 7.3. Suppose Σ2 ⊂ BR ⊂ R3, ∂Σ ⊂ ∂BR, here Σ is a minimalsurface. Then there exists ε > 0 such that if
∫Σ|A|2 < ε, then
(R− |x|)2|A|2(x) ≤ 1
What’s more, |A|2(0) ≤ 1R2
The key idea is to find a scaling invariant quantity. Note the condition∫Σ|A|2 < ε itself is a scaling invariant quantity in dimension 2.
20
Proof. Let us define F (x) = (R − |x|)2|A|2(x). Fix p ∈ Σ with F (p) =maxx∈BR
F (x). Since F vanishes on the boundary, p must in the interior of Σ.
We want to show F (p) ≤ 1. Show by contradiction: suppose F (p) > 1. Defineσ > 0 such that σ2|A|2(p) = 1
Next we define Σ to be the rescaled surface which is Σ rescale by 1σ around
p. Then Σ is still minimal. After the rescaling we have:
supB1(p)∩Σ
|A|2 ≤ 4|A|2(p) = 4|A|2(p)σ2 = 1
i.e.|A|2 ≤ 1 on this rescaling ball. So by Simons inequality we have
∆|A|2 ≥ −2|A|2|A|2 ≥ −2|A|2
Then we applied mean value property we get:
|A|2(p) ≤ e
π
∫B1(p)∩Σ
|A|2 =e
π
∫Bσ(p)∩Σ
|A|2
So if we choose ε ≤ π4e we will get a contradiction. Thus we finish the proof.
There is a similar curvature estimate of minimal graph by Heinz, which isthe following theorem:
Theorem 7.4. (Heinz) Suppose Σ is a minimal graph in R3 over DR ⊂ R2,then
|A|2(point above 0) ≤ C
R2
21
The proof of the theorem uses Choi-Schoen estimate once if we have a cur-vature integral
∫|A|2 bound. To do this, we need to notice that graph is stable
with quadratic area growth, hence we can apply stability equation.
remark 7.1. It is interesting to ask the question: why the condition in Choi-Schoen is a bound of
∫BR∩Σ
|A|2?There are several ways to see it is a natural bound in minimal surface theory.
First thing is the stability inequality. Suppose our minimal surface is stable, thenwe can get a bound of
∫BR∩Σ
|A|2. Another thing is Gauss-Bonnet formula,
by Gauss-Bonnet we can also get a bound of∫BR∩Σ
|A|2, notice we use Gaussequation here.
This is one reason why minimal surface in higher dimension is more difficult,because
∫BR∩Σ
|A|2 is no longer a scaling invariant in higher dimension, but the
quantity we can get from stability inequality is still∫BR∩Σ
|A|2.
8 Curvature Estimate for Intrinsic Ball
When we study a submanifold, we always need to aware that the intrinsic ge-ometry and the extrinsic geometry of the submanifold may be different. Sosometimes we also need intrinsic estimate for the minimal surface.
From now on we will use Br(p) to denote the extrinsic ball, and Br(p) todenote the intrinsic ball on Σ, i.e. Br(p) = x ∈ Σ|distΣ(x, p) ≤ r.
We need several lemmas to proceed the curvature estimate for intrinsic balls.First of all, we want a locally graphic lemma:
lemma 8.1. Suppose |A| ≤ 1 on B1(p), then we have B 12(p) is a graph over TpΣ
with the gradient |∇u| ≤ 1, here we assume u is the graph function. Moreover,∂B 1
2(p) ∩B 1
4(p) = ∅
This fact can be viewed intuitively: suppose the second fundamental doesnot change a lot near some point, then in a small neighborhood of that pointthe normal vector will also does not change a lot, so locally it looks like a graph.
Proof. Let d(t) = distS2(n(p), n(γ(t)), if we can show this distance is no morethan π
4 then it is locally a graph. Note d is a Lipschitz function, hence we cantake differentiation:
|d′(t)| ≤ |∇γ′n| ≤ |A| ≤ 1
Hence we know d(t) ≤ t ≤ 12 for t ≤ 1
2 . So B 12(p) is locally a graph with
gradient no more than 1.Next we show that ∂B 1
2(p) ∩ B 1
4(p) = ∅. To show this, we only need to
prove that any geodesic γ starting from p will leave B 14(p) after time 1
2 . Set
f(t) = 〈γ(t), γ′(0)〉. Then we only need to show f( 12 ) > 1
4 , which means thisgeodesic at least in γ′(0) direction can go out of the ball.
Note f(0) = 0, and take derivative:
22
f ′(t) = 〈γ′(t), γ′(0)〉
f ′′(t) = 〈∇γ′γ′, γ′(0)〉 = 〈A(γ′, γ′)n, γ′(0)〉
so f ′(0) = 1, |f ′′| ≤ 1. By integration we have f(t) ≥ t− 12 t
2, hence we getthe result.
The next theorem concerning the estimate from Gauss-Bonnet theorem. Re-call
Theorem 8.1. (Gauss-Bonnet) Suppose Σ is a two dimensional surface withpossible boundary ∂Σ which is piecewise smooth. Then we have:∫
Σ
K +
∫∂Σ
kg +∑
angles = 2πχ(Σ)
Here K is the Gauss curvature, kg is the geodesic curvature and χ is theEuler characteristic.
From Gauss-Bonnet theorem, we can get calculate the area of surfaces. Letl(r) = |∂Br(p)| and Ar = |Br(p)|. Then take derivative we have
A′(r) = l(r), l′(r) =
∫∂Br
kg = 2π −∫BrK
Now assume Σ is minimal, then we have K = − 12 |A|
2, and then
l′(r) = 2π +1
2
∫Br|A|2
Then take integration we have
l(r) = 2πr +1
2
∫ r
0
∫Bs|A|2ds
A(r) = πr2 +1
2
∫ r
0
∫ t
0
∫Bs|A|2ds
Corollary 8.1. Suppose Σ is a minimal disk and BR ∩ ∂Σ = ∅. Then we have
2(A(R)− πR2) =
∫ R
0
∫ t
0
∫Bs|A|2 =
1
2
∫BR|A|2(R− r)2
Here r is the intrinsic distance to the center of the ball.
To prove the corollary, we only need to do integration by part of two functionsf(t) =
∫ t0
∫Bs |A|
2ds and g(t) = 12 (R− t)2.
By this formula we can derive the following area bound:
Theorem 8.2. (Colding-Minicozzi) Suppose Σ is a stable two sided minimaldisk, BR ∩ ∂Σ = ∅, then we have A(R) ≤ 4π
3 R2
23
Proof. We use R− r as a cut-off function in stability inequality, then we have∫BR
(R− r)2|A|2 ≤∫BR|∇(R− r)|2
Since the left hand side is 4A(R)− 4πR2 by previous formula, and the righthand side is A(R), we get the result.
Corollary 8.2. Let Σ a stable 2 sided complete minimal surface in R3, then Σis flat.
Only note from previous theorem if Σ is a disk, then it has quadratic areagrowth, which means it is parabolic. Hence it is flat. If Σ is not a disk, considerits universal cover, which is still stable, then from disk case we can get theresult.
Similar to this result, Schoen conjectured that: If Σ3 ⊂ R4 is a complete2-sided stable minimal hypersurface, then it is flat. In 1975, Schoen-Simon-Yauproved a special case: when Area(BR ∩ Σ) ≤ CR3, then the conjecture holds.
8.1 Colding-Minicozzi Estimate for Embedded MinimalDisk
Theorem 8.3. Given constant C1, there exists C2 such that if Σ is an embeddedminimal disk in R3, then ifmathcalB2s ⊂ Σ−∂Σ and
∫B2s|A|2 ≤ C1, we have sup
Bs|A|2 ≤ C2
s2 , here C2 only
depends on C1.
Note this theorem implies the classical Bernstein theorem for embeddingminimal surface with integration bounded for curvature.
Before prove the theorem, we prove a key lemma.
lemma 8.2. Given C0 > 0 there exists a constant ε > 0 such that if Σ is aminimal disk, B9r ⊂ Σ − ∂Σ and
∫B9r|A|2 ≤ C, and
∫B9r−Br |A|
2 ≤ ε, then we
have supBr|A|2 ≤ C
r2 (In fact, we can show supBr|A|2 ≤ 1
r2 )
Before we prove this lemma, let us see what’s the difference between thiskey lemma and the Choi-Schoen estimate. For Choi-Schoen, we need a globaltiny bound for the integration of curvature. Here we only need an upper boundof integration of curvature in the whole space, and only a tiny bound in theannulus around the ball we want to estimate.
From a certain perspective, this means if the area around the ball hasbounded curvature integration, then the ball it self can not have pointwise largecurvature.
Proof. Let Q = B8r − B2r. Then for any p ∈ Q, we have∫Br(p)
|A|2 ≤ ε. So we
can apply Choi-Schoen estimate in this ball, hence
24
|A|2(p) ≤ Cε
r2
For any p ∈ Q. Also, our previous locally graphic lemma can applied tothis small ball. Now we show that Q is locally a graph over a plane and theprojection down will cover the annulus D3r −D2r of the plane.
Now fix p ∈ ∂B2s, and let TpΣ is horizontal. Then given any q ∈ Q, we canjoin p, q by a path in Σ with length at most 6r + 1
2 |∂B2r|. To construct such apath, we only first connected q to ∂B2r by a segment with length no more than6r, then we use the path on the boundary to connected that point to p.
By the length formula before, we have
l(R) = 2πR+1
2
∫ R
0
∫Bt|A|2dt
So we can estimate
l(2r) ≤ 4πr + r
∫B2r
|A|2 ≤ (4π + C0)r
Then we get a path connected p, q with length lies in Q no more than C1r.So we have
|n(p)− n(q)| ≤ C1r supQ|A| ≤ C1r
C√ε
r= C2
√ε
Then if we take ε small enough we can bound |n(p) − n(q)| as small as wewant. So we proved that Q is locally graphic by previous locally graphic lemma.
Next, we fix x ∈ ∂B2s, and let γx is the geodesic go out of x which isperpendicular to ∂B2s. Then we have
|γ′′x | = |A(γ′x, γ′x)| ≤ C
√ε
r
We integrate to get
|γ′x(0)− γ′x(t)| ≤ C√ε
rt
Integrate 〈γ′x(t), γ′x(0)〉 = 〈γ′x(t)− γ′x(0) + γ′x(0), γ′x(0)〉, we get the estimatethat
〈γx(6r)− γx(0), γ′x(0)〉 > 6r(1− C√ε)
So it means γx(6r) is outside the cylinder over the plane of radius 3r. Thenwe proved that the projection of Q covers the annulus of the plane.
Let’s now consider the cylinder of radius 2r intersects Q, which is a collectionof closed embedded and graphical curves over ∂D2s. Those intersections arecurves, i.e. transversal intersection with Q is because Q is locally a graph (orwe can use Sard type theorem to assert with a different cylinder with a little bit
25
different radius). These curves are closed because of our previous conclusion,because the projection of Q covers D3r −D2r.
Each curve bounds a disk in Σ because Σ is a disk and we have convex hulltheorem in previous sections. Let us choose one such that it bounds Br, we callthis curve σ and the disk it bounds by Γ, Br ⊂ Γ. Since ∂Γ = σ is a graph overthe boundary of a convex domain, by Rado theorem we have Γ is a graph overthe disk D2r
Then we use Heinz theorem we know that
|A|2 ≤ C
r2
on Br. Then we get the result.
By this lemma, we can prove the theorem in the beginning of this section:
Proof. Let us fix ε in the lemma. For given C1, we can choose N large enoughsuch that C1
N ≤ εThen for any p ∈ Bs we consider the annulus Di = B9−i+1s(p) − B9−is(p),
we have
D1 ∪D2 ∪ · · · ∪DN ∪ B9−Ns(p) ⊂ B2s
And they do not intersect each other. So at least one of these componentshave integration curvature no more than ε. So we can apply the lemma and wecan assert that
|A|2(p) ≤ C
9−2N−2s2≤ C
s2
So we finish the proof.
8.2 Schoen-Simon-Yau (SSY) Estimate
In this section we discuss the famous Schoen-Simon-Yau estimate:
Theorem 8.4. Suppose Σn−1 ⊂ Rn is a 2-sided stable minimal hypersurface.
Then for p ∈ [2, 2 +√
2n−1 ] and φ a cut-off function, we have∫
Σ
|A|2pφ2p ≤ Cn,p∫
Σ
|∇φ|2p
One perspective concerning this estimate is the following: from Sobolev em-bedding theorem, we know that (roughly)
∫Σ|∇u|2p can control the oscillation
of u when 2p > n − 1. So if 4 + 2√
2n−1 > n − 1 we can bound the L∞ norm
of A by standard elliptic estimate. So if n = 2, 3, 4, 5, 6, from the SSY estimate
26
we can get a good bound for |A|L∞ , then we can applied Colding-Minicozziestimate in the previous section to get a Bernstein type theorem.
When n ≥ 8 we know Bernstein theorem fails. What about n = 7? We needmore specific analysis to show Bernstein theorem.
f |A|1+2q|∇f ||∇|A|| = (f |A|q|∇|A||)(|A|1+q|∇f |)
So by standard Holder inequality method we have
(2
n− 1− q2 − εq)
∫f2|A|2q|∇|A||2 ≤ (1 +
q
ε)
∫|A|2+2q|∇f |2
If q2 < 2n−1 , we can choose ε small enough such that q2 + εq < 2
n−1Plug into the stability inequality again we have
27
∫|A|4+2qf2 ≤ 2(1 + q)2
∫f2|A|2q|∇|A||2 + 2
∫|A|2+2q|∇f |2
≤ (2 + 2(1 + q)2(q + 1
ε )1
n−1 − q2 − εq)
∫|A|2+2q|∇f |2
Let f = φp for p = 2 + q, we get the result.
Notice that if∫|∇φ|2p → 0 but in a large range φ ≥ 1, we can get Bernstein
theorem. This can be done if vol(Br) ≤ ra for a < 2p, since we can applied thelog cut-off to get
∫|∇φ|2p ≤ (2r)ar−2p
8.3 How to Find Stable Items?
In order to applied the stability inequality, we always want to have some stableguys. There are several ways to find stable minimal items:
(1)Plateau problems(2)Hoffman-Meeks(3)Min-Max(4)Disjoint guys are close to some limit
9 From Helicoid to Multi-graph
The first thing we do to study something is to look at some examples. Besidesplane, the simplest example of embedded minimal surface in R3 is helicoid.
Definition 9.1. Helicoid is the minimal surface in R3 defined by the equation
tan z =y
x
.
Helicoid is a ruled surface, i.e. at each point on the surface there is a straightline through this point complete lies in the surface.
We can compute that on the helicoid, on the z-axis |A| = 1, while in themost of rest part of helicoid, |A| is very tiny. And the whole helicoid spiral alongthe z-axis, turns out to be a multi-graph besides z-axis. This is one propertywe hope to study for general minimal surface.
We want to show the following results:
1. If |A| is large somewhere, then the spiraling happens. i.e. we can findlocal multi-graph;
2. once spiraling happens, the spiraling will keep going.
Roughly speaking, we want to get the following theorem:
28
Theorem 9.1. (Colding-Minicozzi) Let Σ be an embedded minimal disk, Σ ⊂BR, ∂Σ ⊂ ∂BR, and |A| is large in B1. Then Σ contains a multi-value graphin BR/C for C is a fixed constant independent of Σ.
We haven’t define multi-value graph (In later context, multi-graph in abbre-viation), and we also do not explicit say how large |A| is in the unit ball. Wewill say more about them in further context.
9.1 Graph Over a Surface
Image we have a minimal surface Σ, and there is another one Σ′ very close to it,let’s say it looks like a graph over it. Then Σ′ can be viewed ”almost a Jacobifield” over Σ. Since we know positive Jacobi field indicates stability, we hopethis Σ′ can indicate some ”almost stability” of Σ. This is what we are going tostudy in this section.
Suppose Σ is a surface in R3, ei’s are orthonormal frame on Σ. Let n be thenormal vector to Σ.
Now let u be a function on Σ, we construct a graph Σu = p + u(p)n(p) :p ∈ Σ). Note if u ∼= 0, then Σu = Σ.
For a special case u ∼= s is a constant function. Then if γ(t) is a curve in Σ,γ(t) + sn(γ(t)) is a curve in Σs. Note ∇γ′n(γ(t)) = −A(γ′). So we can definea map
B = Id− sA
, which is the differential of the curve in Σs.For general Σu, we have new local frame
ei −→ B(p, u)ei + ui(p)n
Here B(p, u) = Id− u(p)A(p). The new normal vector is
n −→ n−B−1(p, u)(∇u)
Then we can compute the new metric over Σu is
g = B2 +∇u⊗∇u
Then we can further compute the mean curvature
Hu = −∆u− |A|2u+Q
where Q is quadratic in uA,∇u.
9.2 1/2-stability
Let us recall the definition of stability: We say a 2-sided minimal surface Σ isstable if for all φ has compact support,∫
|A|2φ2 ≤∫|∇φ|2
29
Definition 9.2. We call Σ is 1/2-stable if for all φ has compact support,∫|A|2φ2 ≤ 2
∫|∇φ|2
Although it is not as meaningful as stable from geometric view, it is reason-able in the following sense. Recall Corollary 8.1, we get an estimate of area:
4A(R)− 4πR2 =
∫BR|A|2(R− r)2
Suppose the RHS satisfies the estimate∫BR|A|2(R− r)2 ≤ C
∫BR|∇(R− r)|2 = CA(R)
Here if Σ is stable, then the estimate holds for C = 1. Note we can get anarea bound for C < 4. In particular, if we assume 1/2-stable, we get the aboveestimate for C = 2.
So 1/2-stable in fact give us the quadratic area growth for minimal surface,and as a result, we obtain the curvature estimate supBR/∈ |A|
2 ≤ C/R2 for Cindependent of R from Choi-Scheon estimate.
Next lemma will help us to show 1/2-stablity:
lemma 9.1. There is exist a constant δ > 0, such that if Σu is a minimal graphover Σ with u > 0, and |∇u|+ |u||A| < δ, then Σ is 1/2-stable.
Proof. Let w = log u, then plug it into the minimal surface equation for graphsover minimal surface, we get
where a is an operator, b is a vector, c is a function satisfies
|a|, |c| ≤ 3|A||u|+ |∇u||b| ≤ 2|A||∇u|
as long as |u||A|+ |∇u| ≤ δ for δ small.Now suppose φ is any cut-off function on Σ, apply Stokes theorem to div(φ2∇w−
φ2a∇w), we can get 1/2-stability. (Come back if we need more details)
9.3 Analysis of Sectors
Now let us go back to intrinsic geometry of minimal embedded disk Σ. Let r bethe intrinsic distance function to some fixed point in Σ. Note Σ has non positivecurvature, so r is smooth (in fact, the exponential map is a diffeomorphism).
Let Lt = |∂Bt| is the length of boundary of intrinsic ball, then by calculusof variation we have
L′(t) =
∫∂Bt
kg = 2π −∫BtK
30
by Gauss-Bonnet theorem. By minimality, K = 1/2|A|2Suppose σ is a part of ∂Bt. We define the sector St(σ) to be the union of
geodesic go out of normal directions on σ with length ≤ t. You can image it isreally a sector in the plane.
To show it is a well-posed sector, there are two things need to be ruled out.
• some of the geodesics hit σ twice
• two of the geodesics intersect (in particular, one geodesic hit itself)
All these are not hard to be shown by non positive curvature property andGauss-Bonnet theorem, so we omit the proof here.
Now let L(t) be the length of outer boundary of St. Then by co-area formulawe can check
A(t) := Area(St) =
∫ t
0
L(s)ds
and by first variational formula, we have
L′(t) =
∫outer boundary
kg =
∫σ
kg +1
2
∫St
|A|2 = L′(0) +1
2
∫St
|A|2
Integrate both sides
L(t) = L(0) + tL′(0) +1
2
∫ t
0
∫Ss
|A|2ds
Combine all the formula above, we get the formula for the area of sectors
A(R) = RL(0) +1
2R2L′(0) +
1
2
∫ R
0
∫ t
0
∫Ss
|A|2ds
By the same trick as we get area-curvature estimate before (Corollary 8.1),we get the following co-area formula:
A(R) = RL(0) +1
2R2L′(0) +
1
4
∫S(R)
(R− ρ)2|A|2 (1)
Here ρ is the distance function to σ.Sectors play an important role in the proof. They are the small unit where
we want to analyze the behavior of embedded minimal surface.Let us first explain how to use sectors to find multi-graph. Image we have a
lot of sectors, by embeddedness they will not intersect, then we can find manyof them close to each other. If two of them are close to each other, and theyhave bounded |A|, then they will have 1/2-stability. Then we can show they aresuper flat, hence they form some multi-graphs.
